Given

• A body of mass $m$ is moved from the Earth's surface to a height $h = 2R_e$
• Initial distance from Earth's center: $r_1 = R_e$
• Final distance from Earth's center: $r_2 = R_e + h = R_e + 2R_e = 3R_e$

Gravitational potential energy formula The gravitational potential energy of a mass $m$ at distance $r$ from Earth's center is $U = -\frac{GM_em}{r}$

So the increase in potential energy is $\Delta U = U_2 - U_1 = -\frac{GM_em}{r_2} - \left(-\frac{GM_em}{r_1}\right)$ $\Delta U = GM_em\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

Substitute $r_1 = R_e$ and $r_2 = 3R_e$ $\Delta U = GM_em\left(\frac{1}{R_e} - \frac{1}{3R_e}\right)$ $\Delta U = GM_em\left(\frac{2}{3R_e}\right)$ $\Delta U = \frac{2GM_em}{3R_e}$

Use relation between $g$ and $GM_e$ At Earth's surface, $g = \frac{GM_e}{R_e^2}$ Hence, $GM_e = gR_e^2$

Substituting: $\Delta U = \frac{2}{3} \cdot \frac{gR_e^2 m}{R_e}$ $\Delta U = \frac{2}{3}mgR_e$

Match with options $\boxed{\Delta U = \frac{2}{3}mgR_e}$ This corresponds to Option D.